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Question-121817

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Question Number 121817 by shaker last updated on 12/Nov/20
Answered by liberty last updated on 12/Nov/20
lim_(x→1) (((nx^m −n−mx^n +m)/(x^(m+n) −x^n −x^m +1)) ) = lim_(x→1) (((mn(x^(m−1) −x^(n−1) ))/((m+n)x^(m+n−1) −nx^(n−1) −mx^(m−1) )))= lim_(x→1) (((mn((m−1)x^(m−2) −(n−1)x^(n−2) ))/((m+n)(m+n−1)x^(m+n−2) −n(n−1)x^(n−2) −m(m−1)x^(m−2) )) )= ((mn(m−n))/(m^2 +2mn+n^2 −m−n−n^2 +n−m^2 +m)) = ((mn(m−n))/(2mn)) = ((m−n)/2). ▲
limx→1(nxm−n−mxn+mxm+n−xn−xm+1)=limx→1(mn(xm−1−xn−1)(m+n)xm+n−1−nxn−1−mxm−1)=limx→1(mn((m−1)xm−2−(n−1)xn−2)(m+n)(m+n−1)xm+n−2−n(n−1)xn−2−m(m−1)xm−2)=mn(m−n)m2+2mn+n2−m−n−n2+n−m2+m=mn(m−n)2mn=m−n2.▴
Answered by bemath last updated on 12/Nov/20
let x = 1+ w ; w→0 lim_(w→0) ((n/((1+w)^n −1)) − (m/((1+w)^m −1))) = lim_(w→0) ((n/(nw+((n(n−1))/2)w^2 )) − (m/(mw+((m(m−1))/2)w^2 ))) = lim_(w→0) (((2n)/(2nw+n(n−1)w^2 )) − ((2m)/(2mw+m(m−1)w^2 ))) = lim_(w→0) (((2n(2mw+m(m−1)w^2 )−2m(2nw+n(n−1)w^2 ))/((2nw+n(n−1)w^2 )(2mw+m(m−1)w^2 )))) = lim_(w→0) (((4mn+2mn(m−1)w−4mn−2mn(n−1)w)/((2n+n(n−1)w)(2mw+m(m−1)w^2 )))) = lim_(w→0) ((((2mn(m−1)−2mn(n−1))w)/((2n+n(n−1)w)(2m+m(m−1)w)w))) = lim_(w→0) (((2mn(m−1−n+1))/((2n+n(n−1)w)(2m+m(m−1)w))) = lim_(w→0) (((2mn.(m−n))/(2n.2m))) = ((m−n)/2)
letx=1+w;w→0limw→0(n(1+w)n−1−m(1+w)m−1)=limw→0(nnw+n(n−1)2w2−mmw+m(m−1)2w2)=limw→0(2n2nw+n(n−1)w2−2m2mw+m(m−1)w2)=limw→0(2n(2mw+m(m−1)w2)−2m(2nw+n(n−1)w2)(2nw+n(n−1)w2)(2mw+m(m−1)w2))=limw→0(4mn+2mn(m−1)w−4mn−2mn(n−1)w(2n+n(n−1)w)(2mw+m(m−1)w2))=limw→0((2mn(m−1)−2mn(n−1))w(2n+n(n−1)w)(2m+m(m−1)w)w)=limw→0(2mn(m−1−n+1)(2n+n(n−1)w)(2m+m(m−1)w)=limw→0(2mn.(m−n)2n.2m)=m−n2

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